$h(x) = -6x^{2}-3(f(x))$ $g(x) = -5x^{2}+5(f(x))$ $f(t) = -t^{3}+4t-2$ $ h(g(1)) = {?} $
First, let's solve for the value of the inner function, $g(1)$ . Then we'll know what to plug into the outer function. $g(1) = -5(1^{2})+5(f(1))$ To solve for the value of $g$ , we need to solve for the value of $f(1)$ $f(1) = -1^{3}+(4)(1)-2$ $f(1) = 1$ That means $g(1) = -5(1^{2})+(5)(1)$ $g(1) = 0$ Now we know that $g(1) = 0$ . Let's solve for $h(g(1))$ , which is $h(0)$ $h(0) = -6(0^{2})-3(f(0))$ To solve for the value of $h$ , we need to solve for the value of $f(0)$ $f(0) = -0^{3}+(4)(0)-2$ $f(0) = -2$ That means $h(0) = -6(0^{2})+(-3)(-2)$ $h(0) = 6$